1
$\begingroup$

I am having troubles with deriving the upper bounds on option prices. For example, I am trying to determine the upper bound on a European put option.

If I define the put price as $P$, the terminal stock price as $S_T$, the strike as $X$, the risk free rates as $r$, and the time to expiry as $T$, I am able to create the following portfolio;

Short 1 Put, and invest the proceeds.

So at the outset, from the short put position I gain $P$, and I invest an equal amount at the risk free rate, giving me a net position of $0$.

At expiry, if $S_T > X$, I have zero cash flow from the short put position, and have positive cash flow of $P\text{e}^{rT}$. My net position is $P\text{e}^{rT}$, which is a non negative amount.

If however at expiry $S_T < X$, I have a cash flow of $-(X-S_T)$ from the short position and $P\text{e}^{rT}$ from the invested proceeds. This is a net position of $P\text{e}^{rT} - X + S_T$.

From my understanding, for this to satisfy a no arbitrage condition, we need to have $P\text{e}^{rT} - X + S_T < 0$. Rearranging this should show that the upper bound for the put option needs to be $P < X\text{e}^{rT} - S_T\text{e}^{rT}$.

However, when I consult sources (such as Hull), I am told the upper bound is $P < X\text{e}^{rT}$.

Where is the mistake in my work? This will also help me understand the upper bounds for European call options! Thanks!

$\endgroup$
2
$\begingroup$

First, I think you got discounting and compounding wrong. The upper bound for the put price is $P_0 < X e^{-r T}$. This has to hold as long as $S_0 > 0$.

Your approach is generally correct though. Assume that $P_0 = X e^{-r T}$. Then you invest $P_0$ at time $t = 0$ and have $P_0 e^{r T} = X$ at time $t = T$. Now there are three possible states in time $t = T$:

  1. The put option expires worthless. You you have no further obligations and are left with $V_T = X$.
  2. The option expires in-the-money but $S_T > 0$. Then you have to pay out $P_T < X$ from your short put position, leaving you again with a strictly positive cash-flow $V_T > 0$.
  3. We have $S_T = 0$ and thus $P_T = X$. Your net cash-flow is now zero - $V_T = 0$.

As long as $\mathbb{P} \left\{ S_T = 0 \right\} < 1$, this represents a free lottery arbitrage. I.e. you have a zero initial cash-flow $V_0 = 0$, a non-negative terminal cash-flow with probability one $\mathbb{P} \left\{ V_T \geq 0 \right\} = 1$ and a strictly positive terminal cash-flow with a strictly positive probability $\mathbb{P} \left\{ V_T > 0 \right\} > 0$.

But when $\mathbb{P} \left\{ S_T = 0 \right\} = 1$, then any $S_0 \neq 0$ would already represent an arbitrage by itself.

$\endgroup$
-2
$\begingroup$

Have you considered implied volatility ? It might be really high for some volatile stocks, so put price might goes to extremely high values due to iv ...

$\endgroup$
  • 1
    $\begingroup$ This question is about model independent no-arbitrage bounds. I.e. bounds that have to be satisfied in any model an no matter what the unobservable market parameters such as implied volatility are. $P_0 < X e^{-r T}$ holds in the Black-Scholes model even if $\sigma \rightarrow \infty$. $\endgroup$ – LocalVolatility Mar 11 '17 at 19:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.